350. Intersection of Two Arrays II

最新推荐文章于 2022-03-22 11:24:16 发布

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最新推荐文章于 2022-03-22 11:24:16 发布

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原文链接：http://www.cnblogs.com/Reindeer/p/5728923.html

本文介绍了一种计算两个数组交集的算法实现，并提供了详细的代码示例。通过排序和双指针技术来找出两个数组中相同的元素。

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题目：

Given two arrays, write a function to compute their intersection.

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].

Note:

Each element in the result should appear as many times as it shows in both arrays.
The result can be in any order.

Follow up:

- What if the given array is already sorted? How would you optimize your algorithm?
- What if nums1's size is small compared to nums2's size? Which algorithm is better?
- What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?

答案：

思路是先排序再比较

 1 class Solution {
 2 public:
 3     vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
 4         sort(nums1.begin(),nums1.end());
 5         sort(nums2.begin(),nums2.end());
 6         int n1=nums1.size(),n2=nums2.size();
 7         vector<int>s;
 8         int i=0,j=0;
 9         while(i<n1&&j<n2){
10             if(nums1[i]==nums2[j]){
11                 s.push_back(nums1[i]);
12                 i++;
13                 j++;
14             }
15             else if(nums1[i]<nums2[j]){
16                 i++;
17             }
18             else{
19                 j++;
20             }
21         }
22         return s;
23     }
24 };